Question

Помогите найти производную(распишете подробно):
1) y= - 3cosx * (x`2 + 2)
2)y=1/на корень из x
3) y= 1/sinx
4)y=x`4/ 3-x

Answer 1

1)
$y' = (- 3cosx * ( x^{2} +2))' = -3 [(cosx)' * ( x^{2} +2)+cosx * (x^{2} +2)'] = \ \ = -3 [-sinx * ( x^{2} +2)+cosx * 2x] = -3 (2x*cosx-(x^{2} +2)*sinx)$

2) 
$y' = ( frac{1}{ sqrt{x} })' = ( {x}^{-1/2} })' = - frac{1}{2} * x^{-3/2} = - frac{1}{2 sqrt{x^3} }$

3) 
$y' = ( frac{1}{sinx} )' = (sin^{-1}x)' = - 1 *sin^{-2}x*(sinx)' = - frac{cosx}{sin^2x}$

4) 
$y' = ( frac{x^4}{3-x})' = frac{(x^4)'(3-x)-x^4(3-x)'}{(3-x)^2} = frac{4x^3(3-x)-x^4(-1)}{(3-x)^2} = \ \ = frac{12x^3-4x^4+x^4}{(3-x)^2} = frac{12x^3-3x^4}{(3-x)^2} = frac{3x^3(4-x)}{(3-x)^2}$